How to format a pandas dataframe and keep original float precision values-CodePudding

Im using a pandas dataframe to load a received payload, and when format it is not formatted as I would like, example of my code:

import pandas as pd

df = pd.DataFrame([{'A': 2.5e-07, 'B': 2.5e-05, 'C': 2.5e-04, 'D': 0.0001, 'E': 0.01}])

pd.options.display.float_format = '{:f}'.format
print(df.to_string())

pd.set_option('display.float_format', str)
print(df.to_string())

The output:

         A        B        C        D        E
0 0.000000 0.000025 0.000250 0.000100 0.010000

        A       B       C      D    E
0 2.5e-07 2.5e-05 0.00025 0.0001 0.01

what I would like to get:

        A       B       C      D    E
0 0.0000025 0.000025 0.00025 0.0001 0.01

CodePudding user response：

We can use np.format_float_positional:

import numpy as np
import pandas as pd

pd.set_option('display.float_format', np.format_float_positional)

df = pd.DataFrame([
    {'A': 2.5e-07, 'B': 2.5e-05, 'C': 2.5e-04, 'D': 0.0001, 'E': 0.01}
])

print(df.to_string())

Or with an option_context

import numpy as np
import pandas as pd

df = pd.DataFrame([
    {'A': 2.5e-07, 'B': 2.5e-05, 'C': 2.5e-04, 'D': 0.0001, 'E': 0.01}
])

with pd.option_context('display.float_format', np.format_float_positional):
    print(df.to_string())

Both Produce:

           A        B       C      D    E
0 0.00000025 0.000025 0.00025 0.0001 0.01

CodePudding user response：

You can alter the number of floating point decimals but this is applied to all values.

In [61]: df = pd.DataFrame([{'A': 2.5e-07, 'B': 2.5e-05, 'C': 2.5e-04, 'D': 0.0001, 'E': 0.01}])

In [62]: formats = {"A": "{:,.8f}","B": "{:,.6f}", "C": "{:,.5f}", "D": "{:,.4f}", "E": "{:,.2f}"}

In [63]: for col, f in formats.items():
    ...:     df[col] = df[col].map(lambda x: f.format(x))
    ...: 

In [64]: df
Out[64]: 
            A         B        C       D     E
0  0.00000025  0.000025  0.00025  0.0001  0.01

CodePudding user response：

pd.set_option('display.float_format', lambda x:  '%.6f'%x)

CodePudding user response：

Try this:

>>> df.apply(lambda x : ('%.9f' % x).rstrip('0')).to_frame().T

    A           B           C       D       E
0   0.00000025  0.000025    0.00025 0.0001  0.01